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Question 1203086: Hello,
first of all i would like to thank you for reading my question.
I'm struggling with the following problem for days without any remarkable progress: Let V be a finite dimensional Euclidean vector space and f an endomorphism of
V. Let further |det(f)| = 1 and for all v ∈ V hold
∥v∥ ≤ 1 ⇒ ∥f(v)∥ ≤ 1.
Show that f is a linear isometry.
Any Help will be appreciated from the bottom of my heart.
Answer by ikleyn(52803)   (Show Source):
You can put this solution on YOUR website! .
Hello,
first of all i would like to thank you for reading my question.
I'm struggling with the following problem for days without any  noticeable (?) progress:
Let V be a finite dimensional Euclidean vector space and f an endomorphism of
V. Let further |det(f)| = 1 and for all v ∈ V hold
∥v∥ ≤ 1 ⇒ ∥f(v)∥ ≤ 1.
Show that f is a linear isometry.
Any Help will be appreciated from the bottom of my heart.
~~~~~~~~~~~~~~~~~
I will use informal consideration.
Consider action f on the unit orthogonal vectors e(1), e(2), . . . , e(n).
For all these vectors, we have || f(e(k) || <= 1.
If for some k there is strong inequality || f(e(k) || < 1,
then the image of the unit cube will have the volume less than 1.
It means that the determinant of the matrix f will be less than 1,
which contradicts to the condition.
Hence, for all k, || f(e(k) || = 1.
It should imply that f is a linear isometry.
Solved / proved.
You find the way to formalize this informal consideration.
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Notice that your input info (the problem's formulation) is incomplete.
V is not only a linear space. In addition to the structure of a linear space, it has the norm.
So, I suspect, that endomorphism f, which is considered in this problem, conserves
not only the structure of the linear space, but conserves the scalar product of vectors, too.
So, we need to know the whole context, if an extended context does present in the problem.
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